Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $13.8$ years; the standard deviation is $2.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living less than $6.3$ years.
Answer: $13.8$ $11.3$ $16.3$ $8.8$ $18.8$ $6.3$ $21.3$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $13.8$ years. We know the standard deviation is $2.5$ years, so one standard deviation below the mean is $11.3$ years and one standard deviation above the mean is $16.3$ years. Two standard deviations below the mean is $8.8$ years and two standard deviations above the mean is $18.8$ years. Three standard deviations below the mean is $6.3$ years and three standard deviations above the mean is $21.3$ years. We are interested in the probability of a lion living less than $6.3$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lions will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $6.3$ years and the other half $({0.15\%})$ will live longer than $21.3$ years. The probability of a particular lion living less than $6.3$ years is ${0.15\%}$.